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*From*: John Mallinckrodt <ajm@csupomona.edu>*Date*: Sun, 31 Dec 2006 08:41:33 -0800

If a rigid piston has mass, is the force that the gas exerts on the

piston while the piston is accelerating away from the gas less than the

force that the gas would exert on the piston if the piston was at rest?

As John Denker has noted, the case of the massless piston leads to unphysical results. Nevertheless, the mass of the piston is irrelevant for *this* question, which is answered by simply considering the fact that the force exerted on the piston by the gas is the result of collisions between gas molecules and the piston. Thus, if the piston has a velocity directed "*away* from the gas," then the molecules will transfer less momentum to the piston per unit time and the force on it will be *smaller* than it would if it were at rest (all other things being equal) and vice-versa. The acceleration is irrelevant.

To the extent that this effect *matters,* you are in the realm of nonequilibrium thermodynamics which explains your difficulty in finding an expression for the force on the piston.

If one is interested in the long term behavior of the piston, all one can really say is that it will oscillate. If there is no damping and the system is thermally insulated, then the oscillations will continue forever with the total energy (in the spring, in the piston, and in the gas) conserved. The oscillations may be quite irregular because the gas can "slosh around." Indeed, one could imagine seeing some kind of beating phenomenon between the natural frequency of the piston on the spring and that of the gas sloshing back and forth in the container.

If on the other hand there *is* damping (but still no thermal transfer), then the piston will eventually settle at some position that can only be determined by knowing details you are not likely to know about the nonequilibrium thermal behavior of the gas in this particular extremely complicated process.

Finally, if the motion is slow enough that we don't have to worry about the effect of piston motion on the pressure, then we are safely back in the realm of quasistatic equilibrium thermodynamics. The pressure in the gas will obey the quasistatic adiabatic law. Thus, regardless of the form of the damping, the (initially uncompressed) spring will end up compressed by an amount x given by the solution to the equation:

k*x = Po*A*[Vo/(Vo+A*x)]^gamma

where k is the spring constant, Po and Vo are the initial pressure and volume of the gas, A is the piston area and gamma is the usual ratio of specific heats for the gas, this regardless of the nature of the damping force.

John Mallinckrodt

Professor of Physics, Cal Poly Pomona

<http://www.csupomona.edu/~ajm>

and

Lead Guitarist, Out-Laws of Physics

<http://outlawsofphysics.com>

**Follow-Ups**:**Re: [Phys-l] Force on an accelerating piston***From:*John Mallinckrodt <ajm@csupomona.edu>

**References**:**Re: [Phys-l] Force on an accelerating piston***From:*"Dan Crowe" <dcrowe@Loudoun.K12.VA.US>

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